\(\int \frac {(d+e x)^3}{\sqrt {d^2-e^2 x^2}} \, dx\) [828]

Optimal result

Integrand size = 24, antiderivative size = 116 \[ \int \frac {(d+e x)^3}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {5 d^2 \sqrt {d^2-e^2 x^2}}{2 e}-\frac {5 d (d+e x) \sqrt {d^2-e^2 x^2}}{6 e}-\frac {(d+e x)^2 \sqrt {d^2-e^2 x^2}}{3 e}+\frac {5 d^3 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e} \]

[Out]

5/2*d^3*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e-5/2*d^2*(-e^2*x^2+d^2)^(1/2)/e-5/6*d*(e*x+d)*(-e^2*x^2+d^2)^(1/2)/e
-1/3*(e*x+d)^2*(-e^2*x^2+d^2)^(1/2)/e

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {685, 655, 223, 209} \[ \int \frac {(d+e x)^3}{\sqrt {d^2-e^2 x^2}} \, dx=\frac {5 d^3 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e}-\frac {5 d^2 \sqrt {d^2-e^2 x^2}}{2 e}-\frac {5 d (d+e x) \sqrt {d^2-e^2 x^2}}{6 e}-\frac {(d+e x)^2 \sqrt {d^2-e^2 x^2}}{3 e} \]

[In]

Int[(d + e*x)^3/Sqrt[d^2 - e^2*x^2],x]

[Out]

(-5*d^2*Sqrt[d^2 - e^2*x^2])/(2*e) - (5*d*(d + e*x)*Sqrt[d^2 - e^2*x^2])/(6*e) - ((d + e*x)^2*Sqrt[d^2 - e^2*x
^2])/(3*e) + (5*d^3*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(2*e)

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 655

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[e*((a + c*x^2)^(p + 1)/(2*c*(p + 1))),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 685

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*((a + c*x^2)^(p
+ 1)/(c*(m + 2*p + 1))), x] + Dist[2*c*d*((m + p)/(c*(m + 2*p + 1))), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p, x]
, x] /; FreeQ[{a, c, d, e, p}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p
]

Rubi steps \begin{align*} \text {integral}& = -\frac {(d+e x)^2 \sqrt {d^2-e^2 x^2}}{3 e}+\frac {1}{3} (5 d) \int \frac {(d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx \\ & = -\frac {5 d (d+e x) \sqrt {d^2-e^2 x^2}}{6 e}-\frac {(d+e x)^2 \sqrt {d^2-e^2 x^2}}{3 e}+\frac {1}{2} \left (5 d^2\right ) \int \frac {d+e x}{\sqrt {d^2-e^2 x^2}} \, dx \\ & = -\frac {5 d^2 \sqrt {d^2-e^2 x^2}}{2 e}-\frac {5 d (d+e x) \sqrt {d^2-e^2 x^2}}{6 e}-\frac {(d+e x)^2 \sqrt {d^2-e^2 x^2}}{3 e}+\frac {1}{2} \left (5 d^3\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx \\ & = -\frac {5 d^2 \sqrt {d^2-e^2 x^2}}{2 e}-\frac {5 d (d+e x) \sqrt {d^2-e^2 x^2}}{6 e}-\frac {(d+e x)^2 \sqrt {d^2-e^2 x^2}}{3 e}+\frac {1}{2} \left (5 d^3\right ) \text {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right ) \\ & = -\frac {5 d^2 \sqrt {d^2-e^2 x^2}}{2 e}-\frac {5 d (d+e x) \sqrt {d^2-e^2 x^2}}{6 e}-\frac {(d+e x)^2 \sqrt {d^2-e^2 x^2}}{3 e}+\frac {5 d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.72 \[ \int \frac {(d+e x)^3}{\sqrt {d^2-e^2 x^2}} \, dx=\frac {\left (-22 d^2-9 d e x-2 e^2 x^2\right ) \sqrt {d^2-e^2 x^2}}{6 e}-\frac {5 d^3 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{e} \]

[In]

Integrate[(d + e*x)^3/Sqrt[d^2 - e^2*x^2],x]

[Out]

((-22*d^2 - 9*d*e*x - 2*e^2*x^2)*Sqrt[d^2 - e^2*x^2])/(6*e) - (5*d^3*ArcTan[(e*x)/(Sqrt[d^2] - Sqrt[d^2 - e^2*
x^2])])/e

Maple [A] (verified)

Time = 2.48 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.62

method	result	size
risch	\(-\frac {\left (2 x^{2} e^{2}+9 d e x +22 d^{2}\right ) \sqrt {-x^{2} e^{2}+d^{2}}}{6 e}+\frac {5 d^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-x^{2} e^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\)	\(72\)
default	\(\frac {d^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-x^{2} e^{2}+d^{2}}}\right )}{\sqrt {e^{2}}}+e^{3} \left (-\frac {x^{2} \sqrt {-x^{2} e^{2}+d^{2}}}{3 e^{2}}-\frac {2 d^{2} \sqrt {-x^{2} e^{2}+d^{2}}}{3 e^{4}}\right )+3 d \,e^{2} \left (-\frac {x \sqrt {-x^{2} e^{2}+d^{2}}}{2 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-x^{2} e^{2}+d^{2}}}\right )}{2 e^{2} \sqrt {e^{2}}}\right )-\frac {3 d^{2} \sqrt {-x^{2} e^{2}+d^{2}}}{e}\)	\(166\)

[In]

int((e*x+d)^3/(-e^2*x^2+d^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/6*(2*e^2*x^2+9*d*e*x+22*d^2)/e*(-e^2*x^2+d^2)^(1/2)+5/2*d^3/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)
^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.62 \[ \int \frac {(d+e x)^3}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {30 \, d^{3} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (2 \, e^{2} x^{2} + 9 \, d e x + 22 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{6 \, e} \]

[In]

integrate((e*x+d)^3/(-e^2*x^2+d^2)^(1/2),x, algorithm="fricas")

[Out]

-1/6*(30*d^3*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (2*e^2*x^2 + 9*d*e*x + 22*d^2)*sqrt(-e^2*x^2 + d^2))/
e

Sympy [A] (verification not implemented)

Time = 0.52 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.09 \[ \int \frac {(d+e x)^3}{\sqrt {d^2-e^2 x^2}} \, dx=\begin {cases} \frac {5 d^{3} \left (\begin {cases} \frac {\log {\left (- 2 e^{2} x + 2 \sqrt {- e^{2}} \sqrt {d^{2} - e^{2} x^{2}} \right )}}{\sqrt {- e^{2}}} & \text {for}\: d^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- e^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{2} + \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {11 d^{2}}{3 e} - \frac {3 d x}{2} - \frac {e x^{2}}{3}\right ) & \text {for}\: e^{2} \neq 0 \\\frac {\begin {cases} d^{3} x & \text {for}\: e = 0 \\\frac {\left (d + e x\right )^{4}}{4 e} & \text {otherwise} \end {cases}}{\sqrt {d^{2}}} & \text {otherwise} \end {cases} \]

[In]

integrate((e*x+d)**3/(-e**2*x**2+d**2)**(1/2),x)

[Out]

Piecewise((5*d**3*Piecewise((log(-2*e**2*x + 2*sqrt(-e**2)*sqrt(d**2 - e**2*x**2))/sqrt(-e**2), Ne(d**2, 0)),
(x*log(x)/sqrt(-e**2*x**2), True))/2 + sqrt(d**2 - e**2*x**2)*(-11*d**2/(3*e) - 3*d*x/2 - e*x**2/3), Ne(e**2,
0)), (Piecewise((d**3*x, Eq(e, 0)), ((d + e*x)**4/(4*e), True))/sqrt(d**2), True))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.73 \[ \int \frac {(d+e x)^3}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {1}{3} \, \sqrt {-e^{2} x^{2} + d^{2}} e x^{2} + \frac {5 \, d^{3} \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{2 \, \sqrt {e^{2}}} - \frac {3}{2} \, \sqrt {-e^{2} x^{2} + d^{2}} d x - \frac {11 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2}}{3 \, e} \]

[In]

integrate((e*x+d)^3/(-e^2*x^2+d^2)^(1/2),x, algorithm="maxima")

[Out]

-1/3*sqrt(-e^2*x^2 + d^2)*e*x^2 + 5/2*d^3*arcsin(e^2*x/(d*sqrt(e^2)))/sqrt(e^2) - 3/2*sqrt(-e^2*x^2 + d^2)*d*x
- 11/3*sqrt(-e^2*x^2 + d^2)*d^2/e

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.48 \[ \int \frac {(d+e x)^3}{\sqrt {d^2-e^2 x^2}} \, dx=\frac {5 \, d^{3} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{2 \, {\left | e \right |}} - \frac {1}{6} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left ({\left (2 \, e x + 9 \, d\right )} x + \frac {22 \, d^{2}}{e}\right )} \]

[In]

integrate((e*x+d)^3/(-e^2*x^2+d^2)^(1/2),x, algorithm="giac")

[Out]

5/2*d^3*arcsin(e*x/d)*sgn(d)*sgn(e)/abs(e) - 1/6*sqrt(-e^2*x^2 + d^2)*((2*e*x + 9*d)*x + 22*d^2/e)

Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x)^3}{\sqrt {d^2-e^2 x^2}} \, dx=\int \frac {{\left (d+e\,x\right )}^3}{\sqrt {d^2-e^2\,x^2}} \,d x \]

[In]

int((d + e*x)^3/(d^2 - e^2*x^2)^(1/2),x)

[Out]

int((d + e*x)^3/(d^2 - e^2*x^2)^(1/2), x)